Out of 5 apples,10 mangoes,and 15 oranges,15 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The number of ways to distribute $15$ fruits is the coefficient of $x^{15}$ in the expansion of the product of the generating functions for each f

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(A) The number of ways to distribute $15$ fruits is the coefficient of $x^{15}$ in the expansion of the product of the generating functions for each fruit type.For apples: $(1 + x + x^2 + x^3 + x^4 + x^5) = \frac{1-x^6}{1-x}$For mangoes: $(1 + x + x^2 + \dots + x^{10}) = \frac{1-x^{11}}{1-x}$For oranges: $(1 + x + x^2 + \dots + x^{15}) = \frac{1-x^{16}}{1-x}$The generating function is $f(x) = \frac{(1-x^6)(1-x^{11})(1-x^{16})}{(1-x)^3} = (1 - x^6 - x^{11} - x^{16} + x^{17} + x^{21} + x^{22} - x^{33})(1-x)^{-3}$.Using the binomial expansion $(1-x)^{-3} = \sum_{n=0}^{\infty} \binom{n+2}{2} x^n$,we need the coefficient of $x^{15}$:$= \binom{15+2}{2} - \binom{9+2}{2} - \binom{4+2}{2} = \binom{17}{2} - \binom{11}{2} - \binom{6}{2}$$= \frac{17 	imes 16}{2} - \frac{11 	imes 10}{2} - \frac{6 	imes 5}{2} = 136 - 55 - 15 = 66$.Thus,the total number of ways is $66$.

Out of $5$ apples,$10$ mangoes,and $15$ oranges,$15$ fruits are to be distributed between two persons. Find the total number of ways of distribution.

Similar Questions

$(102)^4 = ?$

The expression $[x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$ is a polynomial of degree

If the coefficient of $x^4$ in the expansion of $\frac{x}{(x-1)^2(x-2)}$ is $\frac{m}{n}$ and $|m|, |n|$ are coprime,then $\sqrt{|m+n|}=$

The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3) \ldots (1+x^{100})$ is

$(3+\sqrt{8})^5+(3-\sqrt{8})^5=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module